Vectors

Vectors

A vector has magnitude and direction. In 2D it is written as a column (x, y) or in i, j notation xi + yj (i and j are unit vectors along the axes); in 3D add a k component.

Magnitude and direction

The magnitude (length) of vector a = xi + yj is:

|a| = √(x² + y²)     (3D: √(x² + y² + z²))

A unit vector (magnitude 1) in the direction of a is a ÷ |a|.

Adding and scaling

• Add/subtract vectors component by component.
• Scalar multiple: ka scales the length by k (negative reverses direction).
• Vectors are parallel if one is a scalar multiple of the other.

Position vectors and geometry

• The position vector of a point A is its vector from the origin, OA.
• The vector from A to B is AB = b − a (position vector of B minus A).
• Distance between A and B = |b − a|.
• The midpoint of AB has position vector ½(a + b).

Using vectors in problems

Vectors are used to prove geometric results (e.g. showing lines are parallel or points are collinear) and to describe positions and displacements. Collinear points lie on the same straight line — show that AB and BC are parallel (scalar multiples) sharing point B.

The scalar (dot) product (if in your spec)

a · b = |a||b| cos θ = x₁x₂ + y₁y₂ (+ z₁z₂)

• Used to find the angle between two vectors: cos θ = (a·b) ÷ (|a||b|).
• If a · b = 0, the vectors are perpendicular.

Worked example

Points A(1, 2) and B(4, 6). Find the vector AB and its magnitude.

• AB = b − a = (4 − 1, 6 − 2) = (3, 4).
• |AB| = √(3² + 4²) = √25 = 5. ✓

Common mistakes

• Computing AB as a − b instead of b − a.
• Forgetting to square root when finding magnitude.
• Confusing a position vector (from the origin) with a displacement vector (between two points).

Exam tips

• Use AB = b − a and |b − a| for distances.
• Show vectors are parallel (scalar multiples) to prove collinearity.
• Use the dot product to find angles or test for perpendicularity (a·b = 0).

Key facts to remember

• Vector magnitude |a| = √(x² + y²); unit vector = a/|a|; parallel ⇔ scalar multiple.
• AB = b − a; distance = |b − a|; midpoint = ½(a + b).
• Dot product a·b = |a||b|cos θ = sum of component products; a·b = 0 ⇔ perpendicular.